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| 1. |
(TanA +CosecB)² -(cotB- secA)² =2tanA.CotB(CosecA+SecB) |
| Answer» We have,LHS = (tanA + cosec B)2 - (cotB - sec A)2{tex}\\Rightarrow{/tex}\xa0LHS = (tan2A + cosec2B + 2tanA cosecB ) -\xa0(cot2B + sec2A\xa0- 2cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secABut, Sec2A - tan2A =1 & cosec2A - cot2\xa0A = 1{tex}\\therefore{/tex}\xa0LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA{tex}\\Rightarrow{/tex}\xa0LHS = 2 (tanA cosecB + cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { cosec\\: B } { \\cot B } + \\frac { \\sec A } { \\tan A } \\right){/tex}\xa0[Dividing and multiplying by tanA cotB]{tex}\\Rightarrow{/tex}\xa0LHS = 2tan A cotB{tex}\\left\\{ \\frac { \\frac { 1 } { \\sin B } } { \\frac { \\cos B } { \\sin B } } + \\frac { \\frac { 1 } { \\cos A } } { \\frac { \\sin A } { \\cos A } } \\right\\}{/tex}\xa0[Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]{tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { 1 } { \\cos B } + \\frac { 1 } { \\sin A } \\right){/tex}\xa0= 2tanA cotB ( secB + cosecA ) = RHS. Hence, proved. | |